Помогите решить 2 cos (3x-π/3) - √3=0

2cos (3x-π/3) - √3=0

2cos (3x-π/3) = √3

cos (3x-π/3) = √3/2

3x-π/3=+/-π/6+2πn, n∈Z

3x=+/-π/6+π/3+2πn, n∈Z

x=+/-π/18+π/9+2/3πn, n∈Z

x1=π/18+π/9+2/3πn=π/6+2/3πn, n∈Z

x2 = - π/18+π/9+2/3πn=π/18+2/3πn, n∈Z